Solve the puzzle?

Question

 

ABCDxD=DCBA

Answer 1

D = 1: one times any number will be that number, so if ABCDxD = DCBA, then D has to = 1

Answer 2

Hello Sun,

 

Try 1089 X 9 = 9801.

 

Regards,

Hassan H.

Answer 3

(10^3A + 10^2B + 10C + D)*D = 10^3D + 10^2C + 10B + A

Since AD = D, A = 1.

Since the unit digit of D^2 is 1, D = 9.

Now we have

900B + 90C + 9081 = 100C + 10B + 9001

which can be simplified to

900B + 90C + 80 = 100C + 10B

=> 890B + 80 = 10C

B = 0, C = 8.

 

Answer: 1089*9 = 9801

 

 

Answer 4

1089*9=9801   :)

 

I first tried to think of what digit D could be. Obviously, it couldn't be 0 or 1, but it couldn't be 2 through 8 either, because then A would be at least 4. So D had to be 9 and hence A had to be 1. Then I set up the equation

 

1BC9*9=9CB1

 

(1000+100B+10C+9)*9=9000+100C+10B+1

 

Simplify this and you get

 

89B+8=C

 

Since C<10, B had to be 0, so that C=8.